question_answer

If the radius of the cirumcircle of isosceles triangle ABC is equal to AB=AC, then the angle A is:

A) \[30{}^\circ \]

B) \[60{}^\circ \]

C) \[90{}^\circ \]

D) \[120{}^\circ \]  

Correct Answer: D

Solution :

[d] If the circumradius of triangle ABC be R, then

\[R=\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}\]

where a, b, c has their usual meanings. Given \[\Delta ABC\] is isosceles such that

\[AB=AC\]

Let circumradius be R, then

\[R=\frac{AC}{2\sin B}=AB=AC\Rightarrow \frac{AC}{2\sin B}=AC\]

\[\sin B=\frac{1}{2}\Rightarrow \sin B=\sin \frac{\pi }{6}\Rightarrow \angle B=\frac{\pi }{6}=\angle C\]

We know that \[\angle A+\angle B+\angle C=180{}^\circ =\pi \]

\[\angle A+\frac{\pi }{6}+\frac{\pi }{6}=\pi \Rightarrow \angle A+\frac{\pi }{3}=\pi \]

\[\Rightarrow \angle A=\pi -\frac{\pi }{3}=\frac{2\pi }{3}=\frac{2\times 180}{3}\Rightarrow \angle A=120{}^\circ \]